codeforces 1293A. ConneR and the A.R.C. Markland-N

Problem link

A solution in c++

	#include<bits/stdc++.h>

	using namespace std;

	#pragma GCC optimize("Ofast,no-stack-protector")
	#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
	#pragma GCC optimize("unroll-loops")

	/// Typedef
	typedef long long ll;
	typedef unsigned long long ull;

	#define sc1(a) scanf("%lld",&a)
	#define sc2(a,b) scanf("%lld %lld",&a,&b)

	#define pf1(a) printf("%lld\n",a)
	#define pf2(a,b) printf("%lld %lld\n",a,b)

	#define vpnt(ans) for(ll i = 0; i < ans.size(); i++) cout << ans[i] << (i + 1 < ans.size() ? ' ' : '\n');
	#define apnt(arr, num) for(ll i = 0; i < num; i++) cout << arr[i] << (i + 1 < num ? ' ' : '\n');

	#define mx 200005
	#define mod 1000000007
	#define PI acos(-1.0)
	#define eps 1e-7

	#define size1 1505

	#define no cout << "NO" << endl
	#define yes cout << "YES" << endl
	#define mem(name, value) memset(name, value, sizeof(name))

	#define pb push_back
	#define ff first
	#define ss second
	#define mp make_pair
	#define all(v) (v).begin(), (v).end()
	#define case cout << "Case " << t++ << ": ";

	typedef vector <ll> vll;
	typedef set <ll> sll;
	typedef queue <ll> qll;
	typedef map <ll, ll> mll;
	typedef pair <ll, ll> pll;
	typedef vector <pair <ll , ll> > vpll;

	inline bool cmp(pll a,pll b){ if(a.ff == b.ff)return a.ss < b.ss; return a.ff > b.ff; }

	int main() {

	//freopen("/opt/Coding/clion code/input.txt", "r", stdin);
	//freopen("/opt/Coding/clion code/output.txt", "w", stdout);

	ll n, m, num, tc, t = 1;

	sc1(tc);
	while (tc--)
	{
	sc2(n, m);
	sc1(num);
	sll st;
	ll x;
	for (ll i = 0; i < num; ++i) {
	sc1(x);
	st.insert(x);
	}

	if(st.find(m) == st.end()) {
	cout << 0 << endl;
	continue;
	}

	ll a1 = 0, a2 = 0;

	for (ll i = m; i <= n; ++i) {
	if(st.find(i) == st.end()) {
	a2 = i;
	break;
	}
	}

	for (ll i = m - 1; i > 0; i--) {
	if(st.find(i) == st.end()) {
	a1 = i;
	break;
	}
	}

	if(a1 == 0) {
	cout << abs(m - a2) << endl;
	continue;
	}
	if(a2 == 0) {
	cout << abs(m - a1) << endl;
	continue;
	}

	a1 = abs(m - a1);
	a2 = abs(m - a2);

	ll ans = min(a1, a2);
	pf1(ans);
	}
	}